CHAPTER 8
Parallelograms and Other Polygons
The study of triangles and their relationships provides a tool for investigating other figures: quadrilaterals of various types and polygons with more than four sides. All that you’ve learned about congruence will be used again and again.
Parallelograms
A parallelogram is a quadrilateral with two pairs of opposite sides parallel. If you focus on one pair of parallel lines and view the other lines as transversals, you can show that consecutive interior angles are supplementary:
Consecutive angles of a parallelogram are supplementary.
Drawing a diagonal in a parallelogram divides it into two triangles that can easily be shown to be congruent. By CPCTC, it can be proven that every parallelogram has these properties:
Opposite sides of a parallelogram are congruent.
Opposite angles of a parallelogram are congruent.
If both diagonals are drawn, proving congruent triangles and showing corresponding parts are congruent will let you show:
The diagonals of a parallelogram bisect each other.
To prove that a quadrilateral is a parallelogram, you need to prove one of the following:
Both pairs of opposite sides are parallel.
Both pairs of opposite sides are congruent.
One pair of opposite sides is both parallel and congruent.
Both pairs of opposite angles are congruent.
Diagonals bisect one another.
The area of a parallelogram is equal to the length of the base times the height, or perpendicular distance between the parallel sides. (A = bh)
EXAMPLE
has AB = CD = 12 cm and the perpendicular distance between 
and 
is 5 cm, the area of is A = bh = 12 ⋅ 5 = 60 square centimeters.
Apply properties of parallelograms to solve these questions.
1. ABCD is a parallelogram with diagonal 
. M is the midpoint of and N is the midpoint of . Prove 
is parallel to 
.
2. The diagonals of parallelogram ABCD intersect at point E. AE = x + 2, BE = y − 1, CE = 3x − 1, and DE = 2y − 3. Find the length of diagonal and diagonal 
.
3. PQRS is a parallelogram. m∠Q = 11x + 13 and m∠S= 15x − 23. Find the measure of ∠P.
4. In the figure below, ∠BAC ≅ ∠ECD, ∠CED ≅ ∠ECD, and 
. Prove ABCD is a parallelogram.
5. In the figure below, 
, ∠AFB ≅ CGD, and ∠BAE ≅ ∠DCE. Prove ABCD is a parallelogram.
6. Find the area of a parallelogram with perimeter of 26 cm if the short side is x cm, the long side is x + 3 cm, and the altitude drawn to the long side is x − 1 cm.
7. In parallelogram WXYZ, the area is calculated using the longer side and the altitude drawn to the longer side. If the shorter side is half the longer side, how does the altitude drawn to the shorter side compare to the altitude to the longer side?
8. ABCD, EFGH, and EICJ are parallelograms. Prove ∠A ≅ ∠G.
Rectangles, Rhombuses, and Squares
A rectangle is a parallelogram with a right angle. Because every rectangle is a parallelogram, it has all the properties of a parallelogram. Because opposite angles are congruent and consecutive angles are supplementary, all four angles in a rectangle are right angles. It is also possible to show that the diagonals of a rectangle are congruent.
EXAMPLE
In rectangle ABCD, AB = CD = 4 and AD = BC = 8. Because all the angles of the rectangle are right angles, the Pythagorean theorem can be applied. The diagonals BD = AC = 
.
A rhombus is a parallelogram with four sides of equal length. It is a parallelogram, so has all the properties of a parallelogram, including opposite angles congruent and consecutive angles supplementary, but may not have all angles congruent. The diagonals of a rhombus intersect at right angles and bisect the interior angles.
A square is a parallelogram that is both a rectangle and a rhombus. A square has four congruent sides and four right angles. It has all the properties of a parallelogram, as well as the properties of rectangles and rhombuses. Because a square is both equilateral and equiangular, it is a regular polygon.
To prove that a quadrilateral is a rectangle or rhombus or square, you must first prove that it is a parallelogram. Once that is established, you need to prove an additional property to classify the parallelogram further.
Rectangle |
Prove one angle is a right angle, or prove diagonals are congruent. |
Rhombus |
Prove adjacent sides congruent, or prove diagonals are perpendicular or prove the diagonals bisect interior angles. |
Square |
Prove that the parallelogram is both a rectangle and a rhombus. |
The area of a rectangle, like any parallelogram, is A = bh but in a rectangle, the base and height are adjacent sides. The area of a square is also A = bh, but since the square is a rectangle with sides of equal length, that formula becomes A = bh = s ⋅ s = s2. The area of a rhombus can be found as A = bh but can also be found from the lengths of the diagonals. Drawing both diagonals, d1 and d2, creates four right triangles, and each right triangle has legs that measure 
and 
. The area of each of those right triangles is 
. There are four right triangles making up the rhombus, so the area of the rhombus is 
.
EXAMPLE
In a rhombus with diagonals that measure 25 cm and 30 cm, the area is 
square centimeters.
Use properties of quadrilaterals and area formulas to answer the questions below. Be sure to include appropriate units.
1. In quadrilateral ABCD, ABCF and DCBE are parallelograms. 
, and 
. Prove that FBCE is a rectangle.
2. HADE and GHCD are parallelograms. 
and ∠A ≅ ∠C. Prove HBDF is a rhombus.
3. ABCD is a parallelogram, 
and 
. F is the midpoint of 
and E is the midpoint of 
. Prove EBFD is a square.
4. The diagonals of a rhombus measure 42 cm and 56 cm. Find the area of the rhombus.
5. A rhombus with a side of 13 cm has a diagonal of 10 cm. A rectangle with the same altitude as the rhombus has an area of 216 cm2. Find the base of the rectangle to the nearest tenth of a centimeter.
6. If a square and a rhombus have the same perimeter, can they have the same area? Explain your reasoning.
7. ABCE and DCBF are both rhombuses and share side 
. If ∠A ≅ ∠D, explain how you can conclude that ΔBGC is isosceles but not equilateral.
Trapezoids
A trapezoid is a quadrilateral with one pair of parallel sides. The parallel sides, usually referred to as the bases, are of different lengths. The nonparallel sides, sometimes called legs, may be different lengths or the same length. A trapezoid in which the nonparallel sides are congruent is an isosceles trapezoid.
In a trapezoid, the pair of angles at each end of one of the parallel sides is a pair of base angles. In an isosceles trapezoid, a pair of base angles has the same measure. This is not true if the trapezoid is not isosceles. In any trapezoid, the consecutive angles between the parallel sides, along one leg, are supplementary.
The line segment that connects the midpoints of the nonparallel sides of a trapezoid is the midsegment. The midsegment of a trapezoid is parallel to the bases and equal in length to the average of the two bases.
The area of a trapezoid is 
, where b1 and b2 are the lengths of the two parallel sides and h is the perpendicular distance between the parallel sides. You may notice that 
within that formula is the average of the bases, and so the length of the midsegment. The area of a trapezoid is the length of the midsegment times the height.
EXAMPLE
If a trapezoid has a height of 12 cm and bases that measure 8 cm and 14 cm, the length of the midsegment is 
cm and the area of the trapezoid is 11 ⋅ 12 = 132 square centimeters.
Questions in this group of exercises focus on trapezoids. Use the information in the section to solve each exercise. Include appropriate units.
1. In ΔABC, midsegment 
connects M, the midpoint of 
, with N, the midpoint of 
. Prove that quadrilateral AMNC is a trapezoid. If the area of ΔABC is 64 cm2, what is the area of the trapezoid AMNC? Explain your reasoning.
2. ABCD is a parallelogram, and ∠DEC ≅ ∠DCE. Prove quadrilateral ABED is an isosceles trapezoid.
3. ABCD is an isosceles trapezoid, with 
and 
. Diagonals and 
are drawn. Prove 
4. In trapezoid WXYZ, 
. m∠W = 2x + 17, m∠X = 5x + 2, and m∠Y = 4x + 3. Is WXYZ an isosceles trapezoid? Explain your reasoning.
5. In trapezoid ABCD, the midsegment is drawn connecting M, the midpoint of 
, to N, the midpoint of 
. If MBCN is rotated 180°clockwise about N, what type of quadrilateral is formed? Explain how you would find the area of this quadrilateral and show that it leads to the formula for the area of a trapezoid.
6. The key measurements in a trapezoid—the two bases and the height—are known to be 10 cm, 14 cm, and 21 cm, but it is not clear which measurement is the height and which are the bases. Find the area of the trapezoid if (a) h = 10 cm; (b) h = 14; (c) h = 21 cm.
7. A trapezoid has an area of 175 cm2. If the height is 7 cm and one base measures 11 cm, find the length of the other base.
8. Find the height of a trapezoid with bases of 8 in and 12 in, if the area of the trapezoid is 100 in2.
Regular Polygons
A polygon is any closed figure made of line segments that intersect only at their endpoints. A polygon is convex if any diagonal falls inside the polygon. A polygon that is not convex is concave. Concave polygons “cave in,” or have dents in them.
Polygons are named by number of sides:
3 → triangle
4 → quadrilateral
5 → pentagon
6 → hexagon
7 → heptagon
8 → octagon
9 → nonagon
10 → decagon
12 → dodecagon
A convex polygon is regular if it is both equilateral and equiangular. An equilateral triangle is a regular three-sided polygon because it has three congruent sides and three angles of equal size. Remember a rectangle is equiangular but not equilateral, and a rhombus is equilateral but not equiangular. Only a square is a regular quadrilateral.
Using the construction skills you have learned, it is possible to construct several different regular polygons. These constructions inscribe the regular polygon in a circle, creating the polygon with each of its vertices on a circle.
Construct a regular hexagon inscribed in a circle. Draw a radius, connecting the center to any point on the circle. Set the compass opening to the length of the radius of the circle. Place the point of the circle at the point at which the radius meets the circle. Scribe an arc that intersects the circle. Move the point of the compass to the point where the arc crosses the circle, and scribe another arc. Repeat until you return to your starting point. This divides the circle into six congruent arcs. Draw line segments connecting these points in sequence around the circle to create a regular hexagon.
Construct an equilateral triangle inscribed in a circle. Begin as for a regular hexagon, dividing the circle into six congruent arcs. Draw line segments connecting every other point around the circle to create an equilateral triangle.
Construct a square inscribed in a circle. If the center of the circle is marked, draw a line through the center, intersecting the circle in two points, to create a diameter.
If the center is not marked, construct a right angle with its vertex on the circle by first drawing a line that intersects the circle in two places and then constructing a perpendicular to the line at one of the points where it intersects the circle. Connect the points where the sides of this right angle intersect the circle. That produces a diameter.
Whichever method you used to get a diameter, construct the perpendicular bisector of your diameter to create a second diameter, perpendicular to the first. The points at which these diameters intersect the circle divide it into four congruent arcs. Connect those points with line segments to create a square.
These questions focus on properties of polygons and constructions of regular polygons. Use the properties of polygons to answer each question.
1. ABCDEF is a regular hexagon. Prove ΔBDF is equilateral.
2. Identify each figure below. Is it a polygon? Convex or concave? Name it by the number of sides.
3. The figure below is a 12-sided polygon. All its sides are the same length. Is it a regular dodecagon? Explain your reasoning.
4. The construction of a regular hexagon inscribed in a circle divides the circle into six arcs of equal measure. If two arcs of a circle are congruent, the central angles that intercept those arcs are congruent. Explain why, in the diagram below, it is possible to prove 
.
5. The construction of a square inscribed in a circle is based on dividing the circle into four congruent arcs. In the figure below, the circle is divided into four arcs, not all congruent. Is it possible to prove: 
? 
? 
? What type of parallelogram is inscribed?
6. Use a compass and a straightedge to construct an equilateral triangle within a regular hexagon in circle O.
7. Use a compass and a straightedge to construct a square and an octagon in circle O.
Angles of Regular Polygons
Like triangles and quadrilaterals, polygons with more than four sides have as many vertices and interior angles as sides. Each vertex is connected to two other adjacent vertices by sides. By drawing diagonals from a vertex to each of the non-adjacent vertices, you can divide a convex polygon into n − 2 nonoverlapping triangles. The sum of the measures of the interior angles of those triangles is equal to the sum of the measures of the interior angles of the polygon.
In the figure below, m∠A + m∠B + m∠C + m∠D + m∠E = (m∠1 + m∠5) + (m∠2) + (m∠3 + m∠4 + m∠8) + (m∠9) + (m∠6 + m∠7) = (m∠1 + m∠2 + m∠3) + (m∠4 + m∠5 + m∠6) + (m∠7 + m∠8 + m∠9). Each triangle has angles that total 180°, so m∠A + m∠B + m∠C + m∠D + m∠E = 3(180°).
A convex polygon of n sides has n interior angles whose measures add to 180(n − 2)°.
If the polygon is regular, the measure of each interior angle is 
.
If one side of the polygon is extended through a vertex, an exterior angle is formed. Each exterior angle is supplementary to the interior angle at that vertex. If one exterior angle is created at each vertex, the sum of their measures is equal to:
The sum of the exterior angles (one at each vertex) of any convex polygon is 360°.
If the polygon is regular with n sides, each exterior angle measures 
.
A segment drawn from the center of a regular polygon to a vertex is a radius. When radii are drawn to two adjacent vertices, they form a central angle. A regular polygon with n sides has n central angles, and the measure of each is 
. Drawing all n radii divides the regular polygon into n isosceles triangles. Each triangle has a central angle as its vertex angle, and each of its base angles are half of an interior angle of the regular polygon.
Areas of Regular Polygons
The area of a convex polygon with more than four sides can sometimes be found by breaking it into nonoverlapping triangles, finding the area of each, and summing the results. With regular polygons, this can be accomplished by drawing segments, called radii, from the center of the regular polygon to each vertex. In a regular polygon with n sides, this creates n congruent isosceles triangles. If you can find the area of one of these, multiplying by the number of sides will give you the area of the regular polygon.
The formula for the area of a regular polygon is a simplified rule for that process. The area of one triangle is 
, with b equal to one side of the polygon and h equal to the height or altitude of the small isosceles triangle. That height is called the apothem of the regular polygon. The area of the regular polygon with n sides is 
. Because nb is the perimeter of the regular polygon and h is the apothem, that formula can be represented as 
, where a is the length of the apothem and P is the perimeter of the regular polygon.
If you are not given the apothem directly, you may be able to find it from other information. Because the apothem divides the small isosceles triangle into two right triangles, if you know the side and the radius, you can find the apothem using the Pythagorean theorem.
If you don’t have enough information to do that, calculate the measures of the angles in the small isosceles triangle and see if you have a special right triangle and can find the length of the apothem from that. If all that fails, you’ll need to use trigonometry.
Trigonometry comes up later in this book, but if you have a calculator with a key that says tan, you can find the apothem by multiplying half the length of a side times tan(measure of base angle of the little isosceles triangle).
EXAMPLE
In a regular pentagon with a side of 6 cm, the little isosceles triangle has a base angle of 54°, so the apothem is 3tan(54) ≈ 4.13 cm.